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CHEMISTRY 59-320 ANALYTICAL CHEMISTRY Fall - 2010

CHEMISTRY 59-320 ANALYTICAL CHEMISTRY Fall - 2010. Lecture 15 Chapter 10 (continued). Summary of diprotic acid calculations. A diprotic acid H 2 A exists as H 2 A, HA - and A 2- in solution and thus acid may be prepared with any of the above substance.

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CHEMISTRY 59-320 ANALYTICAL CHEMISTRY Fall - 2010

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  1. CHEMISTRY 59-320ANALYTICAL CHEMISTRYFall - 2010 Lecture 15 Chapter 10 (continued)

  2. Summary of diprotic acid calculations • A diprotic acid H2A exists as H2A, HA- and A2- in solution and thus acid may be prepared with any of the above substance. • Case 1: The acid solution is prepared with H2A.

  3. Case 2: The acid solution is prepared with HA-. • Case 3: The solution is prepared with A2-.

  4. 10-2 Diprotic buffers • It acts in the same way as a buffer prepared from a monoprotic acid, except here we have two pairs of acid and conjugate base, i.e. H2A/HA- and HA-/A2-. • Depending on the starting material, one can use one of the following two equations to calculate pH of the buffer solution

  5. Diprotic buffer calculation • 10-12: How many milliliters of 0.202 M NaOH should be added to 25.0 ml of 0.0233 M salicylic acid (2-hydroxybenzoic acid) to adjust the pH to 3.50? • Solution: From Appendix G, pK1= 2.972; pK2 = 13.7 the addition of strong base produced equivalent number of moles of HA-. OH- + H2A ↔ HA- + H2O Assuming x ml of NaOH is required; [HA-] = x*0.202/(0.025+ x) [H2A] = (0.0233*0.025 – x*0.202)/(0.025+x) 3.5 = 2.972 + log([HA-]/[H2A])

  6. 10-3 Polyprotic acids and bases • The treatment of ployprotic acids (HnA) is the same as diprotic acid. • Treating HnA as a monoprotic acid and An- as monobasic. • Other intermediates can be dealt with the following equations for the calculation of pH. • For Hn-1A- • For Hn-2A2-

  7. Example of polyprotic acid calculation • Problem 10-17: (a) calculate the quotient [H3PO4]/[H2PO4-] in 0.0500 M KH2PO4. (b) find the same quotient for 0.0500 M K2HPO4. • Solution: From Appendix G, find pK1 = 2.148, pK2 = 7.198, and pK3 = 12.375. H3PO4↔ H+ + H2PO4- (K1) H2PO4-↔ H+ + HPO42- (K2) HPO42- ↔ H+ + PO43- (K3) (a) using equation 10-13 to calculate [H+], in which F = 0.0500M. The calculation yields [H+] = 1.988 x 10-5 M. then, employing the definition k1 = [H+][H2PO4-]/[H3PO4] to calculate the ratio of [H3PO4]/[H2PO4-] (b) is the same as (a), but uses equation 10-14.

  8. 10-4 Which is the principal species • For monoprotic acid: Following Henderson-Hasselbalch equation pH = pKa + log([A-]/[HA]), when the pH > pKa, the basic species is the dominant one, whereas at pH < pKa, the acidic species is the dominant one. • For polyprotic acid: the reasoning is the same as monoprotic acid, except here there are multiple pKa values to separate different dominant species.

  9. Example of calculating pH and dominant species of polyprotic acid • Problem 10-23: The diprotic acid H2A has pK1 = 4.00 and pK2 = 8.00. (a) At what pH is [H2A] = [HA-]? (b) at what pH is [HA-] = [A2-]? (c) Which is the principle species at pH 2.00: H2A, HA-, A2-? (d) Which is the principle species at pH = 6.00? • Solution: (a) using Henderson-Hasselbalch equation, when [H2A] = [HA-], pH = pKa1, i.e. pH = 4.00. (b) the same as (a), when [H2A] = [HA-], pH = pKa2, pH = 8.00. (c) Since pH = 2.00 < pKa1, the H2A should be the dominant species. (d) since pH > pKa1 & pH < pKa2, HA- is the dominant one.

  10. 10-5 Fractional composition equations • In addition to estimating the dominant species based on pH, one can calculate the fraction of each species. • The fraction of molecules in monoprotic acid

  11. Fractional composition of Diprotic system

  12. Example of calculating pH and dominant species of polyprotic acid Example: calculate the pH and composition of individual solutions of (a) 0.050 0 M H2L+, (b) 0.050 0 M HL, and (c) 0.050 0 ML−. The Acidic Form, H2L+ - diprotic acid

  13. Leucine hydrochloride contains the protonated species H2L+ HL is an even weaker acid, because K2 = 1.80 × 10−10.

  14. 10-6 Isoelectric and isoionic pH • Isoionic pH is the pH of the pure, neutral, polyprotic acid. • Isoelectric pH is the pH at which average charge of the polyorotic acid is 0.

  15. For a diprotic amino acid, the isoelectric pH is halfway between the two pKa values. • For a diprotic amino acid, the isoionic pH is given by equation 10-22, where [H2A-] is not exactly equal to [A-]

  16. Applications of isoelectric pH in biochemistry: Isoelectric focusing

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